Integrand size = 23, antiderivative size = 220 \[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {(3 a-2 b (2+p)) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}-\frac {\left (3 a^2-4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {b \cos ^2(e+f x)}{a+b}\right )}{b^2 f (3+2 p) (5+2 p)}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p} \sin ^2(e+f x)}{b f (5+2 p)} \]
[Out]
Time = 0.16 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3265, 427, 396, 252, 251} \[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=-\frac {\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {b \cos ^2(e+f x)}{a+b}\right )}{b^2 f (2 p+3) (2 p+5)}+\frac {(3 a-2 b (p+2)) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b^2 f (2 p+3) (2 p+5)}-\frac {\sin ^2(e+f x) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b f (2 p+5)} \]
[In]
[Out]
Rule 251
Rule 252
Rule 396
Rule 427
Rule 3265
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \left (1-x^2\right )^2 \left (a+b-b x^2\right )^p \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p} \sin ^2(e+f x)}{b f (5+2 p)}+\frac {\text {Subst}\left (\int \left (a+b-b x^2\right )^p \left (a-2 b (2+p)-(3 a-2 b (2+p)) x^2\right ) \, dx,x,\cos (e+f x)\right )}{b f (5+2 p)} \\ & = \frac {(3 a-2 b (2+p)) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p} \sin ^2(e+f x)}{b f (5+2 p)}-\frac {\left (3 a^2-4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \text {Subst}\left (\int \left (a+b-b x^2\right )^p \, dx,x,\cos (e+f x)\right )}{b^2 f (3+2 p) (5+2 p)} \\ & = \frac {(3 a-2 b (2+p)) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p} \sin ^2(e+f x)}{b f (5+2 p)}-\frac {\left (\left (3 a^2-4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p}\right ) \text {Subst}\left (\int \left (1-\frac {b x^2}{a+b}\right )^p \, dx,x,\cos (e+f x)\right )}{b^2 f (3+2 p) (5+2 p)} \\ & = \frac {(3 a-2 b (2+p)) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}-\frac {\left (3 a^2-4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {b \cos ^2(e+f x)}{a+b}\right )}{b^2 f (3+2 p) (5+2 p)}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p} \sin ^2(e+f x)}{b f (5+2 p)} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.58 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.45 \[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (3,\frac {1}{2},-p,4,\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sqrt {\cos ^2(e+f x)} \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {a+b \sin ^2(e+f x)}{a}\right )^{-p} \tan (e+f x)}{6 f} \]
[In]
[Out]
\[\int \left (\sin ^{5}\left (f x +e \right )\right ) {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{p}d x\]
[In]
[Out]
\[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \]
[In]
[Out]
Timed out. \[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \]
[In]
[Out]
\[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \]
[In]
[Out]
Timed out. \[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int {\sin \left (e+f\,x\right )}^5\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \]
[In]
[Out]